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A 20% ethanol solution R is mixed with ethanol solution S in the propo [#permalink]
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Bunuel wrote:
A 20% ethanol solution R is mixed with ethanol solution S in the proportion 1:3 by volume. This mixture is then mixed with 20% ethanol solution T in the proportion 1:1 by volume. If the resultant mixture is a 31.25% ethanol solution, what is amount of ethanol in solution S ?

A. 46%
B. 48%
C. 50%
D. 52%
E. 55%


Are You Up For the Challenge: 700 Level Questions


let R=volume of solution R
let x=amount of ethanol in solution S
.2R+x*3R+.2*4R=.3125*8R→
x=.5
50%
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Re: A 20% ethanol solution R is mixed with ethanol solution S in the propo [#permalink]
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Bunuel wrote:
A 20% ethanol solution R is mixed with ethanol solution S in the proportion 1:3 by volume. This mixture is then mixed with 20% ethanol solution T in the proportion 1:1 by volume. If the resultant mixture is a 31.25% ethanol solution, what is amount of ethanol in solution S ?

A. 46%
B. 48%
C. 50%
D. 52%
E. 55%


Are You Up For the Challenge: 700 Level Questions


We can let n = the percent of ethanol in solution S and 10 be the number of liters in solution R. Thus the amount of solution S is 30 liters and that of solution T is 10 + 30 = 40 liters.

0.2 x 10 + n/100 x 30 + 0.2 x 40 = 0.3125 x (10 + 30 + 40)

2 + 3n/10 + 8 = 0.3125 x 80

10 + 3n/10 = 25

3n/10 = 15

3n = 150

n = 50

Answer: C
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Re: A 20% ethanol solution R is mixed with ethanol solution S in the propo [#permalink]
Bunuel wrote:
A 20% ethanol solution R is mixed with ethanol solution S in the proportion 1:3 by volume. This mixture is then mixed with 20% ethanol solution T in the proportion 1:1 by volume. If the resultant mixture is a 31.25% ethanol solution, what is amount of ethanol in solution S ?

A. 46%
B. 48%
C. 50%
D. 52%
E. 55%


a1--w1--wavg--w2--a2
w2/w1=a1-wavg/wavg-a2

0.2---1---x---3---y
3/1=0.2-x/x-y

0.2---1---0.3125---1---x
1/1=0.2-0.3125/0.3125-x
x=0.425

3/1=0.2-x/x-y,
3x-3y=0.2-x, 4x-0.2=3y,
3y=4(0.425)-0.2, y=0.5

ans (C)
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Re: A 20% ethanol solution R is mixed with ethanol solution S in the propo [#permalink]
Since the proportions listed are 1:3 and then next quantity added is 1:1 with the former, you get this equation using a weighted average:
The proportions are 1/8, 3/8, and 4/8


.2*\(\frac{1}{8}\) + S*\(\frac{3}{8}\)+.2*\(\frac{4}{8}\)=0.3125

solving gives s=0.5 or 50%

C
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Re: A 20% ethanol solution R is mixed with ethanol solution S in the propo [#permalink]
Let the percentage of concentration of ethanol in the solution S be x.

Since R is mixed with S in the ratio of 1:3,
Let R be 100l and S be 300l.
Total ethanol in the resultant 400l solution = 20 + 3x

Now, this 400l solution is mixed with a 400l solution of T, which is 20% concentrated.

Thus, Total ethanol in the resultant 800l solution = 20 + 3x + 4*20 = 100 + 3x l.....(i)

Since the resultant mixture is a 31.25% concentrated solution,
Total ethanol in the resultant 800l solution = 8*31.25 = 250l.....(ii)

Since (i) = (ii),

100 + 3x = 250,
3x = 150.

Thus, x = 150/3 = 50.

Thus, the correct option is C.
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Re: A 20% ethanol solution R is mixed with ethanol solution S in the propo [#permalink]
Assume the first mixture is one liter, and then we add another liter of the solution T to get that 1:1 combination, resulting in 2 liters of solution.

So in that one liter we have 1/4 of solution R, which has 20/100 ethanol and 3/4 liter of solution S, which has X/100 ethanol.

We take that one liter solution and add another liter of solution T which has 20% ethanol. The resulting mixture is 2 liters of solution that has 31.25% ethanol.

Weighted average equation gives us:

1/4 * 20/100 + 3/4 * X/100 + 20/100 = 2 * 31.25/100

5 + 3/4X + 20 = 62.5
3/4X = 37.5 -> X=50
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